CalcGospel 國際數學圖譜
返回

IAL 2019 May Q9

A Level / Edexcel / P1

IAL 2019 May Paper · Question 9

题目

Problem

Figure 3 shows a plot of the curve with equation y=sin⁡θy=\sin\theta, 0≤θ≤360∘0\leq\theta\leq360^\circ.

Figure 3

(a) State the coordinates of the minimum point on the curve with equation

y=4sin⁡θ,0≤θ≤360∘.\begin{align*} y=4\sin\theta,\qquad 0\leq\theta\leq360^\circ. \end{align*}
(2)

A copy of Figure 3, called Diagram 1, is shown on the next page.

Diagram 1

(b) On Diagram 1, sketch and label the curves

(i) y=1+sin⁡θy=1+\sin\theta, 0≤θ≤360∘0\leq\theta\leq360^\circ,

(ii) y=tan⁡θy=\tan\theta, 0≤θ≤360∘0\leq\theta\leq360^\circ.

(2)

(c) Hence find the number of solutions of the equation

(i) tan⁡θ=1+sin⁡θ\tan\theta=1+\sin\theta that lie in the region 0≤θ≤2160∘0\leq\theta\leq2160^\circ,

(ii) tan⁡θ=1+sin⁡θ\tan\theta=1+\sin\theta that lie in the region 0≤θ≤1980∘0\leq\theta\leq1980^\circ.

(3)

解答

(a)

解法一

思路

展开

sin⁡θ\sin\theta 在 270∘270^\circ 处取最小值 −1-1。乘以 44 后，最小值变成 −4-4。

答题过程

展开

The minimum point is

(270∘,−4).\begin{align*} (270^\circ,-4). \end{align*}

(b)

解法一

思路

展开

y=1+sin⁡θy=1+\sin\theta 是正弦图像上移 11。y=tan⁡θy=\tan\theta 在 90∘90^\circ 和 270∘270^\circ 有竖直渐近线，并经过 (0∘,0)(0^\circ,0)、(180∘,0)(180^\circ,0)、(360∘,0)(360^\circ,0)。

答题过程

展开

For y=1+sin⁡θy=1+\sin\theta, the sketch should pass through

(0∘,1),(90∘,2),(180∘,1),(270∘,0),(360∘,1).\begin{align*} (0^\circ,1),\quad (90^\circ,2),\quad (180^\circ,1),\quad (270^\circ,0),\quad (360^\circ,1). \end{align*}

For y=tan⁡θy=\tan\theta, the sketch should have vertical asymptotes at

θ=90∘,θ=270∘,\begin{align*} \theta=90^\circ,\qquad \theta=270^\circ, \end{align*}

and pass through

(0∘,0),(180∘,0),(360∘,0).\begin{align*} (0^\circ,0),\quad (180^\circ,0),\quad (360^\circ,0). \end{align*}

(c)

解法一

思路

展开

从 0∘0^\circ 到 360∘360^\circ 的图像中，两条曲线有 22 个交点。因为两个函数的周期都是 360∘360^\circ，所以每 360∘360^\circ 重复一次。

答题过程

展开

In one interval of length 360∘360^\circ, there are 22 intersections.

For 0≤θ≤2160∘0\leq\theta\leq2160^\circ,

2160∘=6(360∘),\begin{align*} 2160^\circ=6(360^\circ), \end{align*}

so the number of solutions is

6⋅2=12.\begin{align*} 6\cdot2=12. \end{align*}

For 0≤θ≤1980∘0\leq\theta\leq1980^\circ,

1980∘=5(360∘)+180∘.\begin{align*} 1980^\circ=5(360^\circ)+180^\circ. \end{align*}

The final half-cycle contributes one fewer solution than a full cycle here, so the number of solutions is

11.\begin{align*} 11. \end{align*}